W.Lang, Mar 07 2007
 
 a(n,m) tabl head (triangle) for  A053125


   n\m        0          1         2          3         4         5       6       7    8

   0          1          0         0          0         0         0       0       0    0
   1          4         -2         0          0         0         0       0       0    0
   2         16        -16         3          0         0         0       0       0    0
   3         64        -96        40         -4         0         0       0       0    0
   4        256       -512       336        -80         5         0       0       0    0
   5       1024      -2560      2304       -896       140        -6       0       0    0
   6       4096     -12288     14080      -7680      2016      -224       7       0    0
   7      16384     -57344     79872     -56320     21120     -4032     336      -8    0
   8      65536    -262144    430080    -372736    183040    -50688    7392    -480    9 


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 This triangle  A053125, with offset n=1 and doubled entries, is used for Chebyshev U(2*n-1,x) coefficients 
 (without zeros)and decreasing odd powers of x 
  
 U(2*n-1,x) = sum(U(n,m)*x^(2*(n-m)-1),m=0..n-1), n>=1.

 U(n,m)=((-1)^m)*binomial(2*n-m-1,m)*2^(2*(n-m)-1), n>=m+1>=1, else 0.
 
 U(n,m)=((-1)^m)*sum(binomial(m+p,p)*binomial(2*n,2*(m+p)+1),p=0..n-1-m) 
  
 U(n,m) tabl head (triangle)  
 

   n\m         0          1         2          3         4          5        6       7     8

   1           2          0         0          0         0          0        0       0     0
   2           8         -4         0          0         0          0        0       0     0
   3          32        -32         6          0         0          0        0       0     0
   4         128       -192        80         -8         0          0        0       0     0
   5         512      -1024       672       -160        10          0        0       0     0
   6        2048      -5120      4608      -1792       280        -12        0       0     0
   7        8192     -24576     28160     -15360      4032       -448       14       0     0
   8       32768    -114688    159744    -112640     42240      -8064      672     -16     0
   9      131072    -524288    860160    -745472    366080    -101376    14784    -960    18




 Tis U(n,m) triangle gives therefore the formulae for sin((2*n)*phi)/sin(phi) in terms of odd powers of cos:

 n=1: sin(2*phi)/sin(phi) = 2*cos(phi)

 n=2: sin(4*phi)/sin(phi) = 8*cos(phi)^3 -4*cos(phi)

 n=3: sin(6*phi)/sin(phi) = 32*cos(phi)^5 -32*cos(phi)^3 +6*cos(phi)
 
 .
 .
 .
 

 The scaled triangle u(n,m):= U(n,m)/2^(2*(n-m)-1) is then 

 u(n,m) tabl head (triangle)


    n\m    0      1      2       3       4        5       6       7      8      9

    1      1      0      0       0       0        0       0       0      0      0
    2      1     -2      0       0       0        0       0       0      0      0
    3      1     -4      3       0       0        0       0       0      0      0
    4      1     -6     10      -4       0        0       0       0      0      0
    5      1     -8     21     -20       5        0       0       0      0      0
    6      1    -10     36     -56      35       -6       0       0      0      0
    7      1    -12     55    -120     126      -56       7       0      0      0
    8      1    -14     78    -220     330     -252      84      -8      0      0
    9      1    -16    105    -364     715     -792     462    -120      9      0
   10      1    -18    136    -560    1365    -2002    1716    -792    165    -10
   .
   .
   .


  This coincides with  A053123 which has, however, offset n=0.


  This produces the formulae for sin(2*n*phi)/(2*sin(phi)) in terms of powers of 
  (2^(2*(n-m)-2))*cos(phi)^(2*(n-m)-1)


  n=1: (1/2)*sin(2*phi)/sin(phi) = 1*(2^0)*cos(phi)^1 = cos(phi)

  n=2: (1/2)*sin(4*phi)/sin(phi) = 1*(2^2)*cos(phi)^3 -2*(2^0)*cos(phi)= 4*cos(phi)^3 - 2*cos(phi)

  n=3: (1/2)*sin(6*phi)/sin(phi) = 1*(2^4)*cos(phi)^5 -4*((2^2)*cos(phi)^3) + 3*cos(phi)
  .
  .
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